Question

solve each equation x^5-8x^3+16x=o. factor by grouping?

Answer 1

first take out the x:
x(x^4 - 8x^2 + 16) = 0 so one root is x = 0

now can we factor the part in the parentheses?

Answer 2

x^4 - 8x^2 + 16 = 0

Answer 3

i first multiplied 1 and 16 and found the factors that multiplied to get 16 and add to -5. am i wrong?

Answer 4

-4 and -4
and then i got stuck

Answer 5

- 4 * -4 = 16
and -4x^2 - 4x^2 = -8x^2

so factors are

(x^2 - 4)(x^2 - 4) = 0

Answer 6

can you continue?

Answer 7

where did the x^2 come from

Answer 8

x^2 * x^2 = x^4

Answer 9

wouldn't i take -4 and -4 and plug it into the equation

Answer 10

oh you took the x out first

Answer 11

no
the next step is to
solve x^2 - 4 = 0

Answer 12

i dont normally do that first

Answer 13

i dont normally do it that way

Answer 14

taking it out simplifies the solution

Answer 15

oh

Answer 16

always look out for this - the smaller the powers the easier it is to solve

Answer 17

x^2 - 4 = 0
x^2 = 4
so x = ?

Answer 18

so after taking the x out the equation out the simplified equation is x^4-8x^2=16=0

Answer 19

and then whats next

Answer 20

i've already done whats next you factor to get
(x^2 - 4)(x^2 - 4) = 0
so
(x^2 - 4)= 0
x^2 = 4

so whats the value of x?

Answer 21

2

Answer 22

2 is right but also -2

note there are 2 x^2 - 4 = 0

so we have roots of multiplicity 2 ( double roots)

so our final solution set is

{ 0 , +/- 2, +/- 2}

a total of 5 real roots

Answer 23

* 2 (x^2 - 4)

Answer 24

so the answer is x^2=4; 0, +/-2, +/-2; and r real roots?

Answer 25

5*

Answer 26

the answer is

{0, +/- 2, +/- 2}

Answer 27

thanks

Answer 28

yw

Answer 29

could you give me an example of how to figure out the real zeros in a problem?

Answer 30

positive negative and imaginary

Answer 31

sorry gotta go - its very late in uk

Answer 32

okay im sorry thank you for your help